Question

Question #0cd0c

Answer 1

For a zero-order reaction.

Explanation:

The rate of a reaction is simply a measure of the change in the concentration of the reactants or of the products of a chemical reaction per unit of time.

`"rate" = "change in concentration"/"unit of time"`

For a general `n`-order reaction in which

`overbrace("A" + "A" + ... + "A")^(color(blue)(n)) -> " products"`

you can use the differential rate law to express the relationship between the rate of the reaction, the rate constant `k`, and the concentration of the reactant `["A"]`

`"rate" = ["A"]^color(blue)(n)" " " "color(orange)("(*)")`

The concentration of the reactants is measured in

`["A"] = "mol L"^(-1)`

The rate of the reaction usually expresses the change in concentration of the reactant per second, which means that the units for the rate of the reaction are

`"rate" = "mol L"^(-1) "s"^(-1)`

If you use these units in equation `color(orange)("(*)")`, you will get

`"mol L"^(-1)"s"^(-1) = k * ("mol L"^(-1))^color(blue)(n)`

Rearrange to isolate `k` on one side of the equation

`k = ("mol L"^(-1)"s"^(-1))/("mol L"^(-1))^color(blue)(n)`

You want `k` to have the same units as the rate of the reaction

`k = "mol L"^(-1)"s"^(-1)`

so plug this into the above equation and solve for `color(blue)(n)`

`"mol L"^(-1)"s"^(-1) = ("mol L"^(-1)"s"^(-1))/("mol L"^(-1))^color(blue)(n)`

As you can see, the only way to have a valid equality here is if `color(blue)(n)=0`, since

`("mol L"^(-1))^0 = 1`

This of course implies that you have

`"mol L"^(-1)"s"^(-1) = ("mol L"^(-1)"s"^(-1))/1 = "mol L"^(-1)"s"^(-1)" " " "color(green)(sqrt())`

Therefore, the rate constant of a zero-order reaction has the same units as the rate of the reaction, i.e. `"mol L"^(-1)"s"^(-1)`.